Pchem Kinetic question CTQ 3 and 4 a,b Overall chemical transformation: H_2(g) +

Question

Pchem Kinetic question

CTQ 3 and 4 a,b

Overall chemical transformation: H_2(g) + 2 NO(g) rightarrow N_2O(g) + H_2O(g) Experimentally observed rate law: rate = k(NO)^2(H_2) Proposed mechanism: where k_1 is the rate constant for the bimolecular (forward) step 1, k_-1 is the rate constant for the unimolecular (reverse) step 1, k_2 is the rate constant for the bimolecular step 2.^1 Why is the forward step 1 referred to as a bimolecular step, whereas reverse step 1 is referred to as a unimolecular step? Write an expression (or expressions) which shows how the overall rate of reaction is related to the rate of production or consumption of each of the reactants and products in equation (1). For example, for H_2, rate of reaction = - d(H_2)/dt^1 In principle every step is reversible (as in step 1). However, if the reverse step has an extraordinarily small rate constant it is not explicitly given in the mechanism. Unlike a balanced chemical equation, each elementary step in proposed mechanism describes a specific process which is occurring on a molecular level. Thus, the rate law for any elementary step is completely determined by the reactants in the step. Furthermore, for a bimolecular step the rate is proportional to the number of collisions (only a certain percentage of collisions are successful) and it can be shown that the number of collisions is proportional to the product of the concentrations. Thus, for the forward step 1 min Model 1: rate of forward step 1 = k_1(NO)(NO) = {d(N_2O_2)/dt}_step 1, forward For a unimolecular elementary process the rate depends on the number of molecules. [There will be a certain probability that any one molecule will have sufficient energy in a particular vibrational mode to undergo the decomposition.] Thus, the rate simply depends on the concentration of that molecule. For the reverse step in Model 1, rate of reverse step 1 = k_-1 (N_2O_2) = {- d(N_2O_2)/dt}_step 1, reverse Write an expression for the rate of step 2 in Model 1, and show how it is related to the change of (N_2O_2) with time as in equations (3) and (4). Write expressions for each of the following: a) relationship of rate of forward step 1 to d(NO)/dt b) relationship of rate reverse step 1 to d(NO)/dt

Explanation / Answer

H2 (g) + 2 NO (g) -> N2O (g) + H2O (g)

Rate = k (NO)2 (H2)

1.

In the forward step 1, the rate of reaction is proportional to the no. of collisions of 2 (NO) molecules. Hence, this is a bimolecular step.

In the reverse step 1, the rate of reaction depends on no. of molecules of (N2O2) that has sufficient energy in a particular vibrational mode to undergo decomposition. The rate simply depends on concentration of N2O2, and this is a unimolecular reaction.

2.

By stoichiometry of overall reaction,

Rate of reaction = - d(H2)/dt = - ½ * d(NO)/dt = d(N2O)/dt = d(H2O)/dt

3.

Rate of step 2 = k2 (N2O2) (H2) = {-d(N2O2)/dt}step 2

4.

Rate of forward step 1 = k1 (NO)2 = - ½ * d(NO)/dt

Rate of reverse step 1 = k-1 (N2O2) = ½ * d(NO)/dt

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